There are 
ways of picking 2 of the 10 available positions for a 0. 8 positions remain.
There are 
ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's 
way of doing that.
So we have
The last expression has a more compact form in terms of the so-called multinomial coefficient,
I really hope this is right but 6?
There are two ways you can do this. You can use the longer way:
15+16+17+18+19+20+21+22+23+24 = 195 (starting from the top row, which has 15 logs, up until the bottom row, which has 24 logs, on condition that each second row has one log more than the previous one)
You can also use the formula (where n is the number of rows)
x= (n(first term+last term))/2
x= (10(15+24))/2
x= (10*39)/2
x=390/2
x=195
Either way, there are 195 logs in the stack.
A, x equals twenty-three sevenths
Faucet one fills up
of the bathtub in one minute and faucet two fills up
of the bathtub in one minute. If both are one then they will fill up
So, 1/6 of the bathtub in one minute.
